Interpretation of line integrals with 2 parameterizations. Let $C$ be the curve $x^2+y^2=1, y\ge 0.$ One way to parameterize this curve is $x=\cos \theta,y=\sin \theta.$ Then $$\int_Cdx= \int_{0}^{\pi}1dt=\pi$$ However if we parameterize using $x=t,y=\sqrt{1-t^2}$, $$\int_Cdx=\int_{-1}^{1}\sqrt{1+ \frac{t^2}{1-t^2}}=\arcsin1-\arcsin(-1)=\pi$$
How are both of these answers the same if the orientation of integration is reversed? Is there a geometric explaination for why integrating $\int_{1}^{-1}dx$ gives a negative answer when it should really be the same as the displayed equation?
 A: The value of a line integral over a path is independent of its parameterization, in the following sense: If we integrate over a scalar function, then the previous statement is true without qualification. But if we integrate over a vector field, then two parameterizations with reversed orientations will yield the same value differing by a minus sign. Intuitively, we'd like this accumulation of quantity known as the line integral over a path to be independent of traversal, and depend only on the path itself, i.e., tracing out the curve without regard to "speed"; that's true with regard to scalar functions. But a vector field (e.g., the work done over some curve) will depend on the direction, i.e., orientation, and will differ at most by a minus sign (meaning the absolute values will always be the same). In your specific case, you are integrating over the constant scalar function 1 (i.e., finding the arc length) and so reversing orientation does not matter. 
As it turns out, the general rule, i.e., the case of integrating a vector field $\vec{F}$, can easily be shown and is essentially due to the chain rule. 
More formally, suppose $\vec{x}:[a,b]\to\mathbb{R}^n$ and $\vec{y}:[c,d]\to\mathbb{R}^n$ are two distinct, differentiable, $1-1$ (i.e., injective) parameterizations of the same path. Then, there exists a function $h(t)$ s.t. $\vec{x}(t)=\vec{y}(h(t))$ for every $t$ in the domain of $\vec{x}$. (This correspondence follows since the parameterizations are $1-1$ and map to the same path.) Then we have: $$\int_\vec{x} \vec{F} \cdot d\vec{s}=\int_a^b\vec{F}(\vec{x}(t))\cdot \vec{x}(t)dt=\int_a^b\vec{F}(\vec{y}(h(t))\cdot \vec{y}'(h(t))h'(t)dt $$ 
There's the chain rule! The rest is easy, we use the simple substitution $$\begin{cases}
u=h(t)\\
du=h'(t)dt
\end{cases}
$$ 
and we get $$=\int_c^d\vec{F}(\vec{y}(u))\cdot \vec{y}(u)du=\int_\vec{y} \vec{F} \cdot d\vec{s},$$
so that we see any two arbitrary parameterizations are equivalent. And as I mentioned before, if $\vec{x}$ and $\vec{y}$ have reversed orientation (and again this applies only in the context of vector fields), then we'd have $$\int_\vec{x} \vec{F} \cdot d\vec{s}=-\int_\vec{y} \vec{F} \cdot d\vec{s}$$
 Hopefully this was a good motivation for the result, and sheds light on your problem.
A: Certain things have to be clarified here.
In the first place, $C:=\{(x,y)\ |\ x^2+y^2=1,\ y\geq0\}$ is just a set which happens to be a curve, i.e., a one-dimensional manifold (with boundary!). When we want to consider line integrals along $C$ we have to orient it. On an intuitive level this means that we have to say which of the two points $(\pm 1,0)$ is the initial point of the curve. 
Assume that the parametrization
$$\theta\mapsto \bigl(x(\theta),y(\theta)\bigr):=(\cos\theta,\sin\theta)\qquad (0\leq\theta\leq \pi)$$
produces the intended orientation (i.e., with $(1,0)$ as initial point). Then formally
$dx=x'(\theta)\ d\theta=-\sin\theta\ d\theta$ and therefore 
$$\int_C dx=\int_0^\pi (-\sin\theta)\ d\theta=\cos\theta\biggr|_0^\pi=-1-1=-2\ .$$
The parametrization
$$t\mapsto \bigl(x(t),y(t)\bigr):= (t,\sqrt{1-t^2})\qquad (-1\leq t\leq1)$$
with $x'(t)\equiv1$ corresponds to the opposite orientation of $C$. Therefore
$$\int_C dx=-\int_{-1}^1 x'(t)\ dt=- t\biggr|_{-1}^1=-2\ ,$$
as before.
The simplest way to compute this integral is observing that $dx$ is the differential of the function $(x,y)\mapsto x$. In this way we obtain
$$\int_C dx= x({\rm endpoint})-x({\rm initial\ point})=-1-1=-2\ .$$
